How to calculate max voltage swing in cathode follower?

[tingtong5]

Hello all,

I am currently experimenting with using a tube buffer (simple cathode follower) instead of the opamps based preamp section in my integrated amplifier. Anyway, I was wondering how to calculate the maximum voltage swing (I know gain is slightly less then unity) in the schematic attached.

The reason I'm asking is because I plan on using Vg=-2V (@ 20 mA) and my sources are about 2V rms.
I am using 6N6P tube.

Thanks.

Ronald

[tingtong5]

I do use a grid stopper resistor of 1K2 by the way and also a 56R resistor from cathode to output capacitor. Both are not drawn in the schematic but also irrelevant I guess for calculating max voltage swing.

I plan on using B+=200V, Rl=5K and Rb=100R to get the -2V @ 20 mA operating point of the 6N6P. Rg= 1M.

[DavesNotHere]

here is the formulas that govern you circuit:


common-cathode stage, unbypassed cathode:

Voltage Gain (Output 1):

Av = (mu * Rp)/(Rp + ra + (mu + 1)*Rk)

Input impedance:

Rin = Rg

Output impedance (Output 1):

Rout = [(ra + (mu + 1)*Rk) * Rp] / [(ra + (mu + 1)*Rk) + Rp]

Output impedance (Output 2):

Rout = [(Ra + ra)/(mu + 1) * Rk] / [(Ra + ra)/(mu + 1) + Rk]

Frequency response (Output 1):

f1 = 1/(2*pi*Ci*Rg) - highpass breakpoint due to Ci/Rg f2 = 1/(2*pi*Co*(Rout + Rl)) - highpass breakpoint due to Co/Rout/Rl

Where:

Rg = the grid resistor
Rp = the plate resistor
Rk = the cathode resistor
Rl = the load resistance, or the input resistance of the next stage
Ra = the total load resistance, which is Rp in parallel with the input resistance of the next stage, Rl. If there is no Rl, Ra = Rp.
ra = the internal plate resistance of the tube
mu = the mu of the tube

Note that Rl is ignored in the output impedance calcuations for output 1, not because it doesn't affect output impedance of the overall circuit, but because output impedance is traditionally the impedance of the output of that gain stage looking back into the output. Rl is the input impedance of the following stage, so it is not included. Of course, when calculating overall gain in an amplifier, the loading effect of Rl must be taken into account. In the case of the output taken from the cathode, Rl must be included because it will affect the impedance seen looking back into the cathode.


output 2 is consider "Unity" or no gain.

maximum current is 40 mA. depending on loading, I would say about 4 volts.

[tingtong5]

Thank you Dave, however it still does not answer my question what the maximum (undistorted) voltage output will be...

[Wavebourn]

Originally Posted by tingtong5

Thank you Dave, however it still does not answer my question what the maximum (undistorted) voltage output will be...

Undistorted is zero. You should use plate curves and find an optimal regime according to load impedance.

[tingtong5]

Originally Posted by Wavebourn

Undistorted is zero. You should use plate curves and find an optimal regime according to load impedance.

Sorry, what I ment was the maximum voltage before clipping.

But I think I suddenly do get it... Follow the loadline to Vg=0 and then read the voltage on the x-axis and substract it from the voltage at the operating point, which gives half the voltage swing. Is this correct?

[Wavebourn]

Yes. Loadline will show you both maximal positive and negative swing for given distortions.

[ruffrecords]

Originally Posted by tingtong5

The reason I'm asking is because I plan on using Vg=-2V (@ 20 mA) and my sources are about 2V rms.
I am using 6N6P tube.

Then you may have a problem. 2V rms is 2.828 volts peak. If you are biased only at -2V then your signal's positive peaks are going to take the valve into conduction i.e clipping. To accommodate 2V rms signals without clipping you need to bias at at least -4V.

Cheers

Ian

[Wavebourn]

Ian, you probably forgot that it is about cathode follower: it's bootstrapped, so grid-cathode variations are greatly reduced in respect to input voltage variations. They can be seen from the same plate curves.

[ruffrecords]

Originally Posted by Wavebourn

Ian, you probably forgot that it is about cathode follower: it's bootstrapped, so grid-cathode variations are greatly reduced in respect to input voltage variations. They can be seen from the same plate curves.

Yes, you are absolutely right. I think the schematic from the other poster with a plate load threw me.

In which case, as you say load lines rule the day, with the proviso that the ac load is the important one.

Cheers

Ian